使用Scala可以实现漂亮的XStream序列化吗？

Question

I'm trying out XStream as a way to quickly serialize objects to Xml or JSON to send over the wire, and deserialize. I do want the XML/JSON to be simple/clean.

It seems to work well, I've added a few aliases, but now I've hit a problem, this code:

println(new XStream.toXML(List(1,2,3)))

produces this XML:

<scala.coloncolon serialization="custom">
  <unserializable-parents/>
  <scala.coloncolon>
    <int>1</int>
    <int>2</int>
    <int>3</int>
    <scala.ListSerializeEnd/>
  </scala.coloncolon>
</scala.coloncolon>

I think what is going on is that the Scala List class has its own custom serialization... I wonder if there is a way to override that? I'd prefer to get:

<list>
  <int>1</int>
  <int>2</int>
  <int>3</int>
</list>

Answer 1

The "coloncolon" class, or :: , which is actually called cons , is a subclass of Scala's List . It is used to store the actual elements of a List . The only other List subclass is the class of the singleton object Nil , which represents the empty list.

This is actually doing a reasonable job of serializing it, though it is storing the subclass name -- perhaps a problem when you desserialize it.

I wonder how does it serialize Nil .

Answer 2

I figured out how to write a Converter for Scala's List to get xml as shown above, see:

How can I get XStream to output Scala lists nicely? Can I write a custom converter?